POJ 1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 41222		Accepted: 22338

Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles. 

Write a program to count the number of black tiles which he can reach by repeating the moves described above. 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20. 

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows. 

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 
The end of the input is indicated by a line consisting of two zeros. 

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

Source

Japan 2004 Domestic

思路：水题，bfs即可。

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int map[21][21];
int w,h,sx,sy,ans;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
struct nond{ int x,y; };
int main(){
    while(scanf("%d%d",&w,&h)&&w!=0&&h!=0){
        swap(w,h);
        for(int i=1;i<=w;i++)
            for(int j=1;j<=h;j++){
                char x;cin>>x;
                if(x=='.')    map[i][j]=0;
                else if(x=='#')    map[i][j]=1;
                else if(x=='@'){ sx=i;sy=j;map[i][j]=0; }
            }
        queue<nond>que;ans=1;
        nond tmp;tmp.x=sx;tmp.y=sy;
        que.push(tmp);map[sx][sy]=1;
        while(!que.empty()){
            nond now=que.front();
            que.pop();
            for(int i=0;i<4;i++){
                int cx=dx[i]+now.x;
                int cy=dy[i]+now.y;
                if(cx>=1&&cx<=w&&cy>=1&&cy<=h&&!map[cx][cy]){
                    map[cx][cy]=1;
                    nond mid;mid.x=cx;mid.y=cy;
                    que.push(mid);ans++;
                    //cout<<cx<<" "<<cy<<endl;
                }
            }
        }
        cout<<ans<<endl;
    }
}